Question

Give the value of T1 and T2 in the following figure pulley

Answer 1

Solution:-

Here , net pulling force will be

Weight of 2kg and 2kg block on one side - weight of 2kg block on the other side there , therefore

$\rm \: a = \dfrac{net \: pulling \: force \: }{total \: mass}$

$\rm \: a = \dfrac{(2 \times 10) + (2 \times 10) -( 2 \times 10)}{2 + 2 + 2}$

$\rm \: a = \dfrac{20 + 20 - 20}{6}$

$\rm \: a = \dfrac{20}{6} = \dfrac{10}{2} m {s}^{ - 2}$

For T₁ , let us consider FBD of 2kh block . writing equation of motion , we get

=> T₁ - 20 = 2a

=> T₁ - 20 = 2 × 10/2

=> T₁ - 20 = 10

=> T₁ = 10 + 20

=> T₁ = 30 N

For T₂ , we may consider of FBD of 2 kg block . writing equation of motion, we get

=> 20 - T₂ = 2a

=> 20 - T₂ = 2 × 10/2

=> 20 - T₂ = 10

=> 20 - 10 = T₂

=> 10N = T₂

Answer:- T₁= 30N , T₂ = 10N

Note :- FBD of T₁ and T₂ is given in the form of pictures